How do you integrate $\int \sqrt{9 - x^{2}} (- 2 x) d x$ $intsqrt(9-x^2)(-2x)dx$ ?

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How do you integrate $\int \sqrt{9 - x^{2}} (- 2 x) d x$ $intsqrt(9-x^2)(-2x)dx$ ?

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Answer 1 · 2017-01-21T15:47:31

Let $u = 9 - x^{2}$ $u = 9 - x^2$ . Then, by the power rule, $d u = - 2 x d x$ $du = -2xdx$ and $d x = \frac{d u}{- 2 x}$ $dx = (du)/(-2x)$ .

$= \int \sqrt{u} (- 2 x) \cdot \frac{d u}{- 2 x}$ $=intsqrt(u)(-2x) * (du)/(-2x)$

$= \int \sqrt{u} d u$ $=intsqrt(u)du$

$= u^{\frac{3}{2}} + C$ $=u^(3/2) + C$

$= \frac{2}{3} {(9 - x^{2})}^{\frac{3}{2}} + C$ $=2/3(9 - x^2)^(3/2) + C$

Hopefully this helps!

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How do you integrate $\int \sqrt{9 - x^{2}} (- 2 x) d x$ $intsqrt(9-x^2)(-2x)dx$ ?

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How do you integrate $\int \sqrt{9 - x^{2}} (- 2 x) d x$ $intsqrt(9-x^2)(-2x)dx$ ?